# Is there an intuitive way to understand why $\det(AB)=\det(A)\det(B)\;$?

Let $$A, B \in L(V)$$ (or equivalently $$n \times n$$ matrix), where $$V$$ is a $$n$$-dimensional vector space. There are a multiple of proofs why $$\det(A)\det(B) = \det(AB)$$, but I couldn't find a satisfactory that is intuition-appealing.

First, the most imminent motivation for determinant is volume: $$\det(A)$$ is oriented volume of parallelepiped consisting $$n$$-column vectors of $$A$$. Given three matrices $$A, B, AB$$, we have three distinct parallelepiped each of which has oriented volume $$\det(A)$$, $$\det(B)$$, and $$\det(AB)$$, respectively.

Second, the given $$A, B \in L(V)$$ as in linear transformation, $$AB$$ is simply a composition of linear transformation.

I am trying to relate these two concepts, but it feels that the gap between two concepts are too broad. Is there a nice interpretation that fills the gap between these two concepts ?

• A slightly different question whose answers may answer your question too: math.stackexchange.com/questions/3352492/… Jun 11, 2021 at 7:20
• $\det(A)$ is not a volume. You may treat it as the volume of the parallelepiped whose vertices are the origin and the endpoints of the column vectors of $A$ if you assume that the "unit (hyper)cube" in the current basis has volume $1$, but the determinant is not a volume by nature, because it is dimensionless (i.e. it doesn't carry any measurement unit). As stated in an answer below, it is the relative change of volume (it's a ratio, hence unitless) when an object undergoes the linear transformation represented by $A$. Jun 11, 2021 at 9:49
• the second answer in this link this the proof i prefer for this result. it assumes some basic knowledge of multi linear forms to understand, but i think it gives the most convincing answer. Jun 11, 2021 at 11:33

Here's a slightly modified interpretation of the $$\det(A)$$ as a volume. $$\det(A)$$ is more generally the ratio between volumes in the input and output space.
Thus, take an arbitrary set in the vector space. The determinant of the composition mapping $$AB$$ is the ratio of the volumes of the given set under the transformation. $$B$$ first scales the set's volume by $$\det(B)$$, and then $$A$$ does by $$\det(A)$$, so the total scale factor, which is $$\det(AB)$$ is just the product of these intermediate scale factors. Hence $$\det(AB) = \det(A) \det(B)$$